91Ó°ÊÓ

Parametric equations are a fantastic way to represent curves in the plane using a pair of equations, one for each coordinate. In simple terms, they let us describe a curve by specifying how both the x and y coordinates change with a third variable, often denoted as \( t \), which we can think of as 'time'.

For the problem at hand, we want to find a way to travel along the curve connecting the points \((-1, 2)\) and \((2, 5)\). We assume the curve is a straight line, and the parametric equations for a line turn out to be very straightforward.

We can describe the x-coordinate as \( x(t) = -1 + 3t \) and the y-coordinate as \( y(t) = 2 + 3t \). Here, the parameter \( t \) goes from 0 to 1, which means at \( t = 0 \), we start exactly at point \((-1, 2)\) and at \( t = 1 \), we arrive at point \((2, 5)\). This way, our parametric equations give a perfect trajectory along the curve within the specified bounds.

Reparameterization might sound a bit complex, but it's all about using parametric equations to transform an integral along a curve into an easier-to-handle form.

When we reparameterize an integral, we substitute the expressions for \( x \) and \( y \) derived from the parametric equations and their derivatives with respect to \( t \) into the integral expression. This process turns the original line integral into a single integral with respect to \( t \).

In our problem, we transformed \( dx \) and \( dy \) into \( dt \) terms by finding their derivatives: both \( dx = 3dt \) and \( dy = 3dt \) because our parametric equations \( x(t) = -1 + 3t \) and \( y(t) = 2 + 3t \) show that both x and y change linearly with \( t \).

This reparameterization simplifies our calculations immensely by reducing the problem to one variable (\( t \) instead of two) and allowing us to evaluate a standard definite integral rather than a more complex line integral.

Once the integral has been reparameterized, the next step involves evaluating a polynomial integral. This is much more straightforward because we now have an integral in one variable instead of two, and it specifically simplifies to an integral of a polynomial function.

Our polynomial, in this case, came out to be \( 9(6t^2 + 5t + 2) \). Evaluating this polynomial integral means finding the antiderivative and then calculating the definite integral from 0 to 1.

• The term \( 6t^2 \) becomes \( 2t^3 \)
• \( 5t \) becomes \( \frac{5}{2}t^2 \)
• \( 2 \) becomes \( 2t \)

With these antiderivatives, we apply them over the interval [0, 1], leading to the solution of \( 9(6.5) = 58.5 \).

This is a great example of polynomial integral evaluation simplifying what could be a challenging line integral into simple arithmetic.